How can I show that certain implicit functions can partition $\mathbb R^2$?

Question

Consider the implicit function defined by the equation $$ f(x,y) = 0 \tag{1} \label{eq1} $$ where $f : \mathbb R^2 \rightarrow \mathbb R$ is some continuous function. Suppose that the implicit function defined by \eqref{eq1} partitions $\mathbb R^2$ into two disjoint sets, $\mathbb R^2_+$ and $\mathbb R^2_-$, such that $$ \mathbb R^2 = \mathbb R^2_+ \cup \mathbb R^2_- $$ I've found that, for some examples of $f(x,y)$, I am able to show that for any pair $(x_0,y_0) \in \mathbb R^2$, if $$ f(x_0,y_0) > 0 $$ then $(x_0,y_0)$ belongs to either $\mathbb R^2_+$ or $\mathbb R^2_-$ (but not both), and if $$ f(x_0,y_0) < 0 $$ then $(x_0,y_0)$ belongs to the other disjoint set. The simplest example of this involves the equation of a circle with radius $1$, which is $$ x^2 + y^2 - 1 = 0 $$ If, for any $(x_0,y_0) \in \mathbb R^2$, $$ x_0^2 + y_0^2 - 1 > 0 $$ then $(x_0,y_0)$ is outside the circle, and if $$ x_0^2 + y_0^2 - 1 < 0 $$ then $(x_0,y_0)$ is inside the circle. However, I am unable to prove that this pattern holds more generally for all continuous functions $f$. How can one go about showing this? Does there exist a theorem that addresses this?